Answered: F6 526 N D F7- 873N 2950 mmL6 60° 60°…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Draw the free-body diagram of the given Frame/Truss; (ii). Estimate the support reactions; (iii). Determine the forces in all members indicating the nature of forces; iv). Verify the forces in members AD and AE by another suitable method.

F6 526 N
D
F7- 873N
2950 mmL6
60°
60°
30°
L7
5320mm
Figure T2 (All dimensions are in mm)

Expert Solution

Step 1

free body diagram of truss-

for support reactions

F_x=0 , H_A=0

F_y=0, V_{A +}V_B=526 +873=1399 N

Step 2

B)-

in triangle ADX

$\cos 60 = \frac{A X}{2.95} A X = 1.475 m$

so BX = 3.845m

and triangle ADE is equilateral triangle

so, AD= DE= AE=2.95m

in triangle DEC $\cos 30 = \frac{D C}{D E} \frac{D C}{2.95} = \cos 30 D C = 2.55 m$

in triangle BDX

$\cos 30 = \frac{B X}{B D} \cos 30 = \frac{3.845}{D B} D B = 4.43 m$

in triangle BCY

$\cos 30 = \frac{Y B}{B C} \cos 30 = \frac{Y B}{1.88} Y B = 1.628 m$

taking moment zero at about support B

$V_{A} \times 5.32 - (526 \times 3.845) - (873 \times 1.628) = 0 V_{A} = \frac{3443.714}{5.32} V_{A =} 647.31 N V_{B} = 751.68 N$

Step 3

C)- determine forces in members

at joint A

$F_{A D} \sin 60 + 647.31 = 0 F_{A D} = - \frac{647.31}{\sin 60} F_{A D} = - 747.44 N (c o m p r e s s i o n)$

$F_{A D} \cos 60 + F_{A E} = 0 747.44 \cos 60 = F_{A E} F_{A E} = 373.72 N$ tension

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